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Mathematics Question on Determinants

By using properties of determinants,show that:\begin{bmatrix}1+a^2-b^2& 2ab& -2b\\\ 2ab& 1-a^2+b^2& 2a\\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}$$=(1+a^2+b^2)^3

Answer

△=[1+a2−b22ab−2b 2ab1−a2+b22a 2b−2a1−a2−b2]\triangle=\begin{bmatrix}1+a^2-b^2& 2ab& -2b\\\ 2ab& 1-a^2+b^2& 2a\\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}
Applying R1→R1+bR3R_1 → R_1 + bR_3 and R2→R2−aR3R_2 → R_2 − aR_3, we have:
△=[1+a2+b20−b(1+a2+b2) 01+a2+b2a(1+a2+b2) 2b−2a1−a2−b2]\triangle=\begin{bmatrix}1+a^2+b^2& 0& -b(1+a^2+b^2)\\\ 0& 1+a^2+b^2& a(1+a^2+b^2)\\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}
=(1+a2+b2)2[10−b 01a 2b−2a1−a2−b2]=(1+a^2+b^2)^2\begin{bmatrix}1&0&-b\\\ 0&1&a\\\ 2b& -2a& 1-a^2-b^2\end{bmatrix}
Expanding along R1R_1, we have:
△=(1+a2+b2)2[(1)[1a−2a1−a2−b2]−b[01 2b−2a]]△=(1+a^2+b^2)^2\bigg[(1)\begin{bmatrix}1& a \\\\-2a& 1-a^2-b^2\end{bmatrix}-b\begin{bmatrix}0& 1\\\ 2b& -2a\end{bmatrix}\bigg]
=(1+a2+b2)2[1−a2−b2+2a2−b(−2b)]=(1+a^2+b^2)^2[1-a^2-b^2+2a^2-b(-2b)]
=(1+a2+b2)2(1+a2+b2)=(1+a^2+b^2)^2(1+a^2+b^2)
=(1+a2+b2)3=(1+a^2+b^2)^3